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Svetach [21]
3 years ago
9

Can you help me please to solve this 105x+9.50=x+14.74

Mathematics
2 answers:
-BARSIC- [3]3 years ago
8 0
The answer to the question

kap26 [50]3 years ago
6 0

x  = 0.05038461

hope this helps

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Is he correct and show work please
mario62 [17]

9514 1404 393

Answer:

  yes

Step-by-step explanation:

f(0) = -3 +0.5×0 = -3

f(1) = -3 +0.5×1 = -2.5 = f(0) +0.5

Two linear functions are the same if they agree at two points. Here, f(0) and f(1) are the same for both forms of the function, so these different forms represent the same relation.

Diego is correct.

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3 years ago
A right rectangular prism is shown. The base rectangle has side lengths of 10 centimeters and 3 centimeters. The height of the p
Fiesta28 [93]

B 21 cm2 is the answer good luck

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4 years ago
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The perpendicular bisector of the line segment connecting the points $(-3,8)$ and $(-5,4)$ has an equation of the form $y = mx +
docker41 [41]

Answer:

m = -1/2 and b = 6.5

Step-by-step explanation:

To find the slope of the original line segment, we have to do the change in y/the change in x:

(4-8)/(-5--3) = -4/-2 = 2

2 is the slope of the original line segment, but since this is the perpendicular bisector, we have to take the negative reciprocal of 2 so m = -1/2

To find b we substitute the values of x, y, and m into the equation. Let's use the x value of -3 and the y value of 8:

y = mx + b

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8 = 3/2 + b

6.5 = b

5 0
4 years ago
What is a number line for 5a + 18 < -27?
Colt1911 [192]

Step-by-step explanation:

5a + 18 <  - 27 \\  \\  \therefore \: 5a  <  - 27 - 18 \\  \\  \therefore \: 5a  <  - 45 \\  \\  \therefore \: a  <   \frac{ - 45}{5}  \\  \\ \therefore \: a  <    - 9

8 0
3 years ago
Gravel is being dumped from a conveyor belt at a rate of 15 ft3/min, and its coarseness is such that it forms a pile in the shap
gayaneshka [121]

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0.13 ft/min

Step-by-step explanation:

We are given that

\frac{dV}{dt}=15ft^3/min

We have to find the increasing rate of change of height of pile  when the pile is 12 ft high.

Let d be the diameter of pile

Height of pile=h

d=h

Radius of pile,r=\frac{d}{2}=\frac{h}{2}

Volume of pile=\frac{1}{3}\pi r^2 h=\frac{1}{12}\pi h^3

\frac{dV}{dt}=\frac{1}{4}\pi h^2\frac{dh}{dt}

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\frac{dh}{dt}=\frac{15\times 4}{\pi(12)^2}

\frac{dh}{dt}=0.13ft/min

7 0
4 years ago
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